Prove quadratic equation (y=ax2+bx+c) has only one line of symmetry

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arrebusnaavbz · Accepted Answer

Step 1Symmetry around x=s is expressed byax2+bx+c=a(2s−x)2+b(2s−x)+cfor all x.By identification of the powers of x, we get the compatibility conditionb+2as=0ors=−b2aStep 2Consider an arbitrary parabola and an arbitrary line. By rotation we can bring the line on the axis y. Then the equation of the rotated parabolaax2+2bxy+cy2+2dx+2ey+f=0must be invariant to a change of the sign of x and it must reduce toax2+cy2+2ey+f=0.For this equation to represent a parabola, we must have c=0. Then,ax2+2ey+f=0is a parabola for which the axis of symmetry is y. Hence any line which is an axis of symmetry is the standard axis.Step 3Note that in the case ac≠0 (centered conic), by translation we can further reduce toax2+cy2+f=0and the line is one of the known axis of symmetry. Hence there are exactly two of them.

Prove quadratic equation \(\displaystyle{\left({y}={a}{x}^{{{2}}}+{b}{x}+{c}\right)}\) has only one

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